L2 Recovery of Integrable Functions and Consequences
Corollary 1.2 (Parseval’s Identity)
Let $f$ be integrable function, and $a_n= \hat{f}(n)$. Then $\lim_{N\to\infty}\sum_{n=-N}^N\lvert a_n\rvert^2$ converges to $\lVert f \rVert^2$.
<Proof>
By Pythagorean theorem,
\[\begin{align*} \lVert f \rVert^2 & = \lVert f-S_N(f)\rVert^2 + \lVert S_N(f)\rVert^2 \\ & = \lVert f-S_N(f)\rVert^2 + \frac{1}{2\pi} \int_{-\pi}^{\pi}\left( \sum_{n=-N}^N a_n e^{inx}\right)\left(\overline{\sum_{m=-N}^N a_m e^{imx}}\right) dx \\ &=\lVert f-S_N(f)\rVert^2 + \frac{1}{2\pi} \int_{-\pi}^\pi \sum_{n=-N}^N\sum_{m=-N}^N a_n\overline{a_m}e^{i(n-m)x} dx \\ &=\lVert f-S_N(f)\rVert^2 + \sum_{n=-N}^N\sum_{m=-N}^N a_n\overline{a_m}\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(n-m)x}dx\right) \\ &=\lVert f-S_N(f)\rVert^2 + \sum_{n=-N}^N \lvert a_n\rvert^2, \end{align*}\]where the last equality follows from
\[\begin{align*} \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(n-m)x}dx = \begin{cases} 1 &\text{if } n=m,\\ 0 &\text{if } n\neq m. \end{cases} \end{align*}\]Thus, the cross terms vanish only after integration.
We know that $\lim_{N\to\infty}\lVert f-S_N(f)\rVert^2=0$. Thus,
\[\begin{align*} \lim_{N\to\infty}\sum_{n=-N}^N \lvert a_n\rvert^2&=\lVert f\rVert^2 - \lim_{N\to\infty}\lVert f-S_N(f)\rVert^2 \\ &= \lVert f\rVert^2. \end{align*}\]$\therefore \lVert a_n\rVert_{\ell^2}= \lVert f\rVert_{L^2}$.
\[\tag*{$\square$}\]Theorem (Polarized Parseval’s identity)
Let $f$ and $g$ be integrable on the circle, with Fourier coefficients ${a_n}$ and ${b_n}$, respectively. Then
\[\begin{align*} \frac{1}{2\pi} \int_0^{2\pi} f(\theta)\overline{g(\theta)}d\theta = \sum_{n=-\infty}^\infty a_n \overline{b_n}. \end{align*}\]<Proof>
In any Hermitian inner product space, one has the polarization identity,
\[\begin{align*} \langle f, g \rangle = \frac{1}{4}\left(\lVert f+g\rVert^2 - \lVert f-g\rVert^2 + i(\lVert f+ig \rVert^2-\lVert f-ig\rVert^2)\right). \end{align*}\]Using this in $L^2$ and Parseval’s identity, we get
\[\begin{align*} \langle f, g\rangle_{L^2} &= \frac{1}{4}\left( \sum \lvert a_n +b_n\rvert^2 - \sum \lvert a_n-b_n\rvert^2 + i(\sum\lvert a_n +ib_n\rvert^2 -\sum\lvert a_n-ib_n\rvert^2) \right) \\ &=\frac{1}{4}\left(\lVert \{a_n + b_n\}\rVert^2-\lVert \{a_n-b_n\}\rVert^2 +i (\lVert \{a_n+ib_n\}\rVert^2-\lVert \{a_n-ib_n\}\rVert^2) \right) \\ &=\langle \{a_n\}, \{b_n\}\rangle_{\ell^2}. \end{align*}\] \[\tag*{$\square$}\]Theorem 1.3 (Riemann-Lebesgue Lemma)
Since $f$ is integrable on the compact interval $[0,2\pi]$, it is bounded and hence $f\in L^2$. By Parseval’s identity,
\[\begin{align*} \sum_{n=-\infty}^{\infty}\lvert \hat{f}(n)\rvert^2=\lVert f\rVert_{L^2}^2 <\infty. \end{align*}\]Therefore, the series $\sum_{n=-\infty}^{\infty}\lvert \hat{f}(n)\rvert^2$ converges. In particular, its terms must converge to $0$. That is,
\[\begin{align*} \lvert \hat{f}(n)\rvert^2 \to 0 \quad \text{as } \lvert n\rvert\to\infty. \end{align*}\]Taking square roots,
\[\begin{align*} \lvert \hat{f}(n)\rvert \to 0 \quad \text{as } \lvert n\rvert\to\infty. \end{align*}\]Therefore,
\[\begin{align*} \hat{f}(n)\to 0 \quad \text{as } \lvert n\rvert\to\infty. \end{align*}\]Thus, in our setting, the Riemann-Lebesgue lemma follows directly from Parseval’s identity. The proof below gives a direct argument using integrability.
An equivalent reformulation of this proposition is that if $f$ is integrable on $[0,2\pi]$, then
\[\begin{align*} &\int_0^{2\pi} f(\theta)\sin(n\theta) d\theta \to 0 \quad \text{as } \lvert n\rvert\to\infty \\ &\int_0^{2\pi} f(\theta)\cos(n\theta) d\theta \to 0 \quad \text{as } \lvert n\rvert\to\infty. \end{align*}\]<Proof> First suppose that $f$ is real-valued. Let $\epsilon >0$ be given. Since $f$ is integrable, there is a partition $\mathcal{P}={ 0=x_0, \ldots, 2\pi=x_l}$ of $[0,2\pi]$ such that
\[\begin{align*} \left\lvert \int_0^{2\pi}f(x)dx - L(f, \mathcal{P}) \right\rvert < \frac{\epsilon}{2}. \end{align*}\]Now define a step function $\varphi: [0,2\pi]\to\mathbb{R}$ such that
\[\begin{align*} \varphi(x)= \begin{cases} c_i=\inf_{[x_{i-1}, x_i]}f(x) \quad &\text{if } x \in (x_{i-1}, x_i) \\ f(x_i) \quad &\text{if } x \in \{x_0, \ldots, x_l\} \end{cases} \end{align*}\]Then $\int_0^{2\pi}\varphi(x)dx = \sum_{i=1}^l c_i(x_i - x_{i-1}) = L(\mathcal{P},f)$. Since $\varphi(x)\leq f(x)$ except at finitely many partition points,
\[\begin{align*} \int_0^{2\pi}\lvert f(x)-\varphi(x)\rvert dx =\int_0^{2\pi}f(x)dx-\int_0^{2\pi}\varphi(x)dx < \frac{\epsilon}{2}. \end{align*}\]By triangle inequality,
\[\begin{align*} \left\lvert \int_0^{2\pi} f(x)\cos(nx) dx\right\rvert & \leq \left\lvert \int_0^{2\pi}(f(x)-\varphi(x))\cos(nx) dx\right\rvert + \left\lvert \int_0^{2\pi} \varphi(x)\cos(nx) dx\right\rvert \\ &\leq \int_0^{2\pi} \lvert f(x)-\varphi(x)\rvert dx + \sum_{i=1}^l \frac{\lvert c_i\rvert}{\lvert n\rvert} \lvert \sin(nx_{i})-\sin(nx_{i-1}) \rvert \\ &<\frac{\epsilon}{2} + \frac{2}{\lvert n\rvert}\sum_{i=1}^l\lvert c_i\rvert. \end{align*}\]Since $\sum_{i=1}^l\lvert c_i\rvert$ is fixed, for sufficiently large $\lvert n\rvert$,
\[\begin{align*} \frac{2}{\lvert n\rvert}\sum_{i=1}^l\lvert c_i\rvert<\frac{\epsilon}{2}. \end{align*}\]Therefore,
\[\begin{align*} \left\lvert \int_0^{2\pi} f(x)\cos(nx) dx\right\rvert<\epsilon. \end{align*}\]$\therefore \lim_{\lvert n \rvert \to \infty} \int_0^{2\pi}f(x)\cos(nx)dx=0$. Similarly we can show that
\[\begin{align*} &\int_0^{2\pi} f(\theta)\sin(n\theta) d\theta \to 0 \quad \text{as } \lvert n\rvert\to\infty \\ &\int_0^{2\pi} f(x)e^{-inx}dx \to 0 \quad \text{as } \lvert n\rvert\to\infty. \end{align*}\]For complex-valued $f$, apply the result separately to the real and imaginary parts of $f$.
\[\tag*{$\square$}\]Lemma 1.4
Let $f$ be a bounded function on the compact interval $[a,b]$. If $c\in (a,b)$ and if for small $\delta>0$ the function $f$ is integrable on the intervals $[a,c-\delta]$ and $[c+\delta, b]$, then $f$ is integrable on $[a,b]$.
<Proof>
Suppose that $\lvert f(x) \rvert \leq M$ for all $x\in [a,b]$. Let $\epsilon>0$ be given and take $\delta>0$ such that $4\delta M < \epsilon/3$. Then there is a partition $\mathcal{P}_1$ of $[a,c-\delta]$ and partition $\mathcal{P}_2$ of $[c+\delta, b]$ such that
\[\begin{align*} &U(f, \mathcal{P}_1) -L(f, \mathcal{P}_1) < \frac{\epsilon}{3} \\ &U(f, \mathcal{P}_2)- L(f, \mathcal{P}_2) < \frac{\epsilon}{3} \end{align*}\]since $f$ is integrable on $[a, c-\delta]$ and $[c+\delta, b]$, respectively. Take a partition $\mathcal{P}_3={c-\delta, c+\delta }$ of $[c-\delta, c+\delta]$. Then
\[\begin{align*} \sup_{x\in [c-\delta, c+\delta]} f(x) &\leq M \\ \inf_{x\in [c-\delta, c+\delta]}f(x) &\geq -M. \end{align*}\]Thus,
\[\begin{align*} U(f, \mathcal{P}_3) - L(f, \mathcal{P}_3) \leq 4 \delta M < \frac{\epsilon}{3}. \end{align*}\]Let $\mathcal{P} = \bigcup_{i=1}^3 \mathcal{P}_i$. Then $U(f, \mathcal{P})- L(f,\mathcal{P}) < \epsilon$.
$\therefore f$ is integrable on $[a,b]$.
\[\tag*{$\square$}\]Remark
Convergence in $L^2$ does not guarantee that the Fourier series converges for any $\theta$.
Theorem 2.1
Let $f$ be an integrable function on the circle which is differentiable at $\theta_0$. Then $\lim_{N\to\infty}S_N(f)(\theta_0) = f(\theta_0)$.
<Proof>
Define
\[\begin{align*} F(t) = \begin{cases} \frac{f(\theta_0-t)-f(\theta_0)}{t} \quad &\text{ if } t \neq 0 \text{ and } \lvert t \rvert < \pi \\ -f^\prime(\theta_0) & \text{ if } t= 0 \end{cases} \end{align*}\]$F$ is bounded near $0$ since $f$ is differentiable there. For a small $\delta >0$, the function $F$ is integrable on $[-\pi, -\delta] \cup [\delta, \pi]$ since $f$ is integrable. By lemma 1.4, $F$ is integrable on $[-\pi, \pi]$.
Let $D_N(x)= \sum_{n=-N}^N e^{inx}$ be Dirichlet kernel and recall that $S_N(f)(x) = (f*D_N)(x)$. Then
\[\begin{align*} S_N(f)(\theta_0) - f(\theta_0) &= \frac{1}{2\pi} \int_{-\pi}^\pi f(\theta_0-t)D_N(t) dt - f(\theta_0) \\ &=\frac{1}{2\pi}\int_{-\pi}^\pi (f(\theta_0-t)-f(\theta_0))D_N(t) dt \quad (\because \frac{1}{2\pi}\int_{-\pi}^\pi D_N(t)dt=1) \\ &=\frac{1}{2\pi} \int_{-\pi}^\pi F(t)tD_N(t)dt \end{align*}\]Recall that
\[\begin{align*} tD_N(t) = \frac{t}{\sin(t/2)}\sin((N+1/2)t) \end{align*}\]and
\[\begin{align*} \sin((N+1/2)t)=\sin(Nt)\cos(t/2) + \cos(Nt)\sin(t/2). \end{align*}\]Thus,
\[\begin{align*} S_N(f)(\theta_0) - f(\theta_0) &= \frac{1}{2\pi} \int_{-\pi}^\pi F(t) \frac{t}{\sin(t/2)}\left( \sin(Nt)\cos(t/2) + \cos(Nt)\sin(t/2)\right) dt \\ &= \frac{1}{2\pi} \int_{-\pi}^\pi F(t)\frac{t\cos(t/2)}{\sin(t/2)} \sin(Nt) dt + \frac{1}{2\pi}\int_{-\pi}^\pi F(t)t\cos(Nt)dt. \end{align*}\]The functions
\[\begin{align*} F(t)\frac{t\cos(t/2)}{\sin(t/2)} \quad\text{and}\quad F(t)t \end{align*}\]are integrable on $[-\pi,\pi]$. Indeed, $\frac{t\cos(t/2)}{\sin(t/2)}$ has a removable singularity at $t=0$ and is bounded on $[-\pi,\pi]$.
Therefore, by the Riemann-Lebesgue lemma,
\[\begin{align*} S_N(f)(\theta_0)-f(\theta_0)\to 0 \end{align*}\]as $N\to\infty$.
$\therefore \lim_{N\to\infty}S_N(f)(\theta_0)=f(\theta_0)$.
\[\tag*{$\square$}\]Corollary 2.2 (Localization principle of Riemann)
Suppose $f$ and $g$ are two integrable functions defined on the circle, and for some $\theta_0$ there exists an open interval $I$ containing $\theta_0$ such that $f(\theta)=g(\theta)$ for all $\theta\in I$. Then $S_N(f)(\theta_0)- S_N(g)(\theta_0) \to 0$ as $N\to \infty$.
The function $f-g$ is 0 on $I$, so it is differentiable at $\theta_0$. By the previous theorem, $S_N(f-g)(\theta_0)\to (f-g)(\theta_0)$ as $N\to\infty$.
$\therefore \lim_{N\to\infty}S_N(f)(\theta_0) - S_N(g)(\theta_0) =0$.
\[\tag*{$\square$}\]Reference
- Elias M. Stein and Rami Shakarchi 『Fourier Analysis: An Introduction』
- Math 139 Fourier Analysis Notes